The differentiable functions $x$ and $y$ are related by the following equation: $y^3-3y^2=11-2x$ Also, $\dfrac{dx}{dt}=4$. Find $\dfrac{dy}{dt}$ when $y=-2$.
Answer: Let's start by differentiating the equation $y^3-3y^2=11-2x$ with respect to $t$. $\begin{aligned} y^3-3y^2&=11-2x \\\\ 3y^2\cdot\dfrac{dy}{dt}-6y\cdot\dfrac{dy}{dt}&=-2\cdot\dfrac{dx}{dt} \\\\ (3y^2-6y)\cdot\dfrac{dy}{dt}&=-2\cdot\dfrac{dx}{dt} \end{aligned}$ We are given that $\dfrac{dx}{dt}=4$, and we want to find $\dfrac{dy}{dt}$ when $y=-2$. Let's plug ${y=-2}$ and ${\dfrac{dx}{dt}=4}$ into the equation we obtained: $\begin{aligned} (3{y}^2-6{y})\cdot\dfrac{dy}{dt}&=-2\cdot{\dfrac{dx}{dt}} \\\\ (3({-2})^2-6({-2}))\cdot\dfrac{dy}{dt}&=-2\cdot{4} \\\\ 24\cdot\dfrac{dy}{dt}&=-8 \\\\ \dfrac{dy}{dt}&=-\dfrac{1}{3} \end{aligned}$ In conclusion, when $y=-2$, the value of $\dfrac{dy}{dt}$ is $-\dfrac{1}{3}$.